Let $A$ be a finitely generated $R$-algebra generated by $x_1,...,x_n$ and $M$ a finitely generated $A$-module. If $x_1,...,x_n\in \sqrt{\operatorname{Ann}_{A}{M}}$, show that $M$ is a finitely generated $R$-module.


We can replace $A$ by $R[x_1,x_2....x_n]/I$ for some ideal $I$ in $R[x_1,x_2....x_n]$. Let $M$ be finitely generated by $m_1,m_2.....m_k$ over $A$. Then for any $m\in M$, there exist $(f_j+I)_{j=1....k}$ in $A$ such that $m=\sum_{j=1}^{k}(f_j+I).m_j$.

Now if $x_1,x_2...,x_n\in {\operatorname{Ann}_{A}(M)}$, then we would be done as then $m=\sum_{j=1}^{k}(c_j.m_j)$, where $c_j$ are constant terms of $f_j$.However I am not able to prove it when we only know that $x_1,x_2...,x_n\in \sqrt{\operatorname{Ann}_{A}{M}}$.

Any help is appreciated.


1 Answer 1


$\exists \ N\in \mathbb N$ such that $x_i^N\in \operatorname{Ann}_AM \ \forall \ i$.

Choose a set of generators $\{m_\alpha\}$ of $M$ over $A$. I claim $M$ is generated over $R$ by $\{x^\beta m_\alpha\}$ where $\beta$ is a multi-index i.e. $\beta=(\beta_1,\dots,\beta_n)\in \mathbb Z_{\geq 0}^n$ such that $|\beta|=\sum_i\beta_i\leq Nn$. Clearly this is a finite set.

Pick any $m\in M$. Write $m=\sum_\alpha c_\alpha m_\alpha$ with $c_\alpha\in A$

Each $c_\alpha$ has a representation of the form $c_\alpha =\sum_\beta r_{\beta,\alpha}x^\beta$ where $r_{\beta,\alpha}\in R$

Now observe that $c_\alpha m_\alpha =\sum_{\beta} r_{\beta,\alpha}x^\beta m_\alpha=\sum_{\beta : |\beta|\leq Nn} r_{\beta,\alpha}x^\beta m_\alpha$ since the other terms will be in the annihilator of $M$.

Thus you see any element is a finite $R$ linear combinations of the generators we chose.

  • $\begingroup$ I didn't think that generating set would be of that type. Thanks! $\endgroup$ Commented May 16, 2020 at 11:01

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